Question

From a standstill an object moves with constant acceleration and covers
5m in the first second. What distance does it travel in the next second?

Answer 1

Answer:

Since the object is at standstill before moving, it means the object starts from rest which implies, u = 0 m/s.

Also it is given that the object covers 5 m in the first second.

Therefore, s = 5 m and t = 1 second.

Using the second equation of motion and the given information we get:

⇒ s = ut + 0.5 × a × t²

⇒ 5 = 0(1) + 0.5 × a × (1)²

⇒ 5 = 0 + 0.5 a

⇒ a = 5 / 0.5

⇒ a = 10 m/s²

Now we are required to calculate the distance traveled between 1st and the 2nd second. Calculating the total distance traveled in 2 seconds we get:

⇒ s = 0(2) + 0.5 × (10) × (2)²

⇒ s = 0 + 20

⇒ s = 20 m

Hence the distance travelled between 1st and the 2nd second is:

⇒ Total Distance travelled in 2 seconds - 5 m

⇒ 20 m - 5 m

⇒ 15 m

Hence the object would travel 15 m within 1st and the 2nd second.

Answer 2

Answer:

Given :-

• From a standstill an object moves with constant acceleration and covers 5 m in first second.

To Find :-

• What is the distance does it travel in the next second.

Formula Used :-

$\clubsuit$ Second Equation Of Motion Formula :

$\mapsto \sf\boxed{\bold{\pink{s =\: ut + \dfrac{1}{2} at^2}}}$

where,

• s = Distance Covered
• u = Initial Velocity
• t = Time Taken
• a = Acceleration

Solution :-

First, we have to find the acceleration :

Given :

• Distance Covered = 5 m
• Initial Velocity = 0 m/s
• Time Taken = 1 seconds

According to the question by using the formula we get,

$\leadsto \bf s =\: ut + \dfrac{1}{2} at^2$

$\leadsto \sf 5 =\: (0)(1) + \dfrac{1}{2} \times a(1)^2$

$\leadsto \sf 5 =\: 0 \times 1 + \dfrac{1}{2} \times a \times (1 \times 1)$

$\leadsto \sf 5 =\: 0 + \dfrac{1}{2} \times a \times 1$

$\leadsto \sf 5 =\: 0 + \dfrac{1}{2} \times a$

$\leadsto \sf 5 =\: \dfrac{0 + 1}{2} \times a$

$\leadsto \sf 5 =\: \dfrac{1}{2} \times a$

$\leadsto \sf \dfrac{5}{\dfrac{1}{2}} =\: a$

$\leadsto \sf \dfrac{5}{1} \times \dfrac{2}{1} =\: a$

$\leadsto \sf \dfrac{10}{1} =\: a$

$\leadsto \sf 10 =\: a$

$\leadsto \sf\bold{\purple{a =\: 10\: m/s^2}}$

Now, again we have to find the distance travelled by an object :

As we know that :

Given :

• Initial Velocity = 0 m/s
• Time Taken = 2 seconds
• Acceleration = 10 m/s²

According to the question by using the formula we get,

$\implies \bf s =\: ut + \dfrac{1}{2} at^2$

$\implies \sf s =\: (0)(2) + \dfrac{1}{2} \times (10)(2)^2$

$\implies \sf s =\: 0 \times 2 + \dfrac{1}{2} \times 10 \times (2 \times 2)$

$\implies \sf s =\: 0 + \dfrac{1}{2} \times 10 \times 4$

$\implies \sf s =\: 0 + \dfrac{1}{2} \times 40$

$\implies \sf s =\: \dfrac{0 + 1}{2} \times 40$

$\implies \sf s =\: \dfrac{1}{\cancel{2}} \times {\cancel{40}}$

$\implies \sf s =\: \dfrac{1}{1} \times 20$

$\implies \sf s =\: \dfrac{20}{1}$

$\implies \sf\bold{\blue{s =\: 20\: m}}$

Now, we have to find the distance travelled by the object in the next second :

$\longrightarrow \sf Total\: Distance\: Travelled =\: 20 - 5$

$\longrightarrow \sf\bold{\red{Total\: Distance\: Travelled =\: 15\: m}}$

${\footnotesize{\bold{\underline{\therefore\: The\: distance\: travelled\: by\: the\: object\: in\: the\: next\: second\: is\: 15\: m\: .}}}}$

▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃

EXTRA INFORMATION :-

❒ Equation Of Motion :

➲ First Equation Of Motion :

$\bigstar\: \: \sf\bold{\pink{v =\: u + at}}\: \: \bigstar$

➲ Second Equation Of Motion Formula :

$\bigstar\: \: \sf\bold{\red{s =\: ut + \dfrac{1}{2} at^2}}\: \: \bigstar$

➲ Third Equation Of Motion Formula :

$\bigstar\: \: \sf\bold{\purple{v^2 =\: u^2 + 2as}}\: \: \bigstar$

where,

• s = Distance Covered
• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• t = Time Taken